Physics Question Over 2 Dimensional Friction Problems

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    Friction Physics
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Angela is trying to determine the speed needed to keep a glass from falling out of a bucket while spinning it in a vertical circle. The discussion covers the forces acting on the glass, including weight and normal force, and clarifies that centripetal force is not an additional force but rather the net inward force required for circular motion. Participants work through the equations needed to find the speed at which the glass will stay in the bucket without the normal force dropping to zero. The conversation also addresses the maximum tension the string can withstand, leading to calculations for the range of speeds that keep the glass secure while preventing the string from breaking. Ultimately, the participants refine their understanding of the forces involved and the correct equations to use for solving the problem.
  • #31
haruspex said:
The glass is blissfully unaware of the tension in the string and of gravity acting on the bucket. Which forces act directly on the glass, and by doing so provide the centripetal force ( so don't answer "centripetal force").

Normal Force?
 
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  • #32
Xazerd said:
Thank You, but I already solved that one now ( I got Net Force= Normal Force- Weight + (Mass*Velocity^2)/radius.)
That equation is not correct, in a couple of ways. Perhaps you've typed it out wrongly?
 
  • #33
pgardn said:
T + mg = (mv^2)/r
Think again. At what point in the loop is the tension greatest?
 
  • #34
haruspex said:
Think again. At what point in the loop is the tension greatest?

Aha again!

I was using the FBD for part a)

The problem said Maximum tension. This would happen at the bottom of the loop.

So T - mg = (mv^2)/r

It always helps to answer the questions that are actually asked.
Sorry to the OP. Thanks to Mr. H!
 
  • #35
pgardn said:
Aha again!

I was using the FBD for part a)

The problem said Maximum tension. This would happen at the bottom of the loop.

So T - mg = (mv^2)/r

It always helps to answer the questions that are actually asked.
Sorry to the OP. Thanks to Mr. H!

So is this the equation I use for part c and d, but I have to combine the mass of both the bucket and glass?
 
  • #36
Xazerd said:
So is this the equation I use for part c and d, but I have to combine the mass of both the bucket and glass?

Not for part c)

You got part c.

Gravity the sole actor for the centripetal force in c, so mass cancels.

Part d) would occur at the bottom of the circular path. So the tension, or more directly the normal force would act up, and gravity acts down. If you combine mass of the glass and the bucket then you can use tension.
 
  • #37
pgardn said:
Not for part c)

You got part c.

Gravity the sole actor for the centripetal force in c, so mass cancels.

Part d) would occur at the bottom of the circular path. So the tension, or more directly the normal force would act up, and gravity acts down. If you combine mass of the glass and the bucket then you can use tension.

Ok. Got it. Thanks!
 
  • #38
Xazerd said:
Ok. Got it. Thanks!

Sorry I did not read the question more carefully and used part a) for everything.
 
  • #39
No Problem!
 

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